Questions: Question 4 The matrix [0.1 0.2; 0.1 0.3] is an input-output matrix for a market where the total output is [10; 25]. Determine the internal production matrix. [6.5; 15.5] [4; 16.5] [6; 8.5] [3.5; 9.5] Question 5 The matrix [0.2 0.3; 0.1 0.1] is an input-output matrix for a market where the total output is [15; 10]. Determine the internal production matrix. [9; 7.5] [4; 5.5] [11; 4.5] [6; 2.5]

Solution Steps

To determine the internal production matrix, we need to solve a system of linear equations. The internal production matrix \( \mathbf{x} \) can be found using the equation \( \mathbf{x} = \mathbf{y} - \mathbf{A} \mathbf{x} \), where \( \mathbf{y} \) is the total output vector and \( \mathbf{A} \) is the input-output matrix. Rearranging gives \( (\mathbf{I} - \mathbf{A}) \mathbf{x} = \mathbf{y} \), where \( \mathbf{I} \) is the identity matrix. We solve for \( \mathbf{x} \) by computing the inverse of \( (\mathbf{I} - \mathbf{A}) \) and multiplying it by \( \mathbf{y} \).

For Question 4, the input-output matrix \( \mathbf{A} \) is given by: \[ \mathbf{A} = \begin{bmatrix} 0.1 & 0.2 \\ 0.1 & 0.3 \end{bmatrix} \] and the total output vector \( \mathbf{y} \) is: \[ \mathbf{y} = \begin{bmatrix} 10 \\ 25 \end{bmatrix} \]

We need to find the internal production matrix \( \mathbf{x} \) using the equation: \[ \mathbf{x} = (\mathbf{I} - \mathbf{A})^{-1} \mathbf{y} \] where \( \mathbf{I} \) is the identity matrix: \[ \mathbf{I} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \] Calculating \( \mathbf{I} - \mathbf{A} \): \[ \mathbf{I} - \mathbf{A} = \begin{bmatrix} 1 - 0.1 & 0 - 0.2 \\ 0 - 0.1 & 1 - 0.3 \end{bmatrix} = \begin{bmatrix} 0.9 & -0.2 \\ -0.1 & 0.7 \end{bmatrix} \] The inverse of this matrix is computed, yielding: \[ (\mathbf{I} - \mathbf{A})^{-1} \approx \begin{bmatrix} 1.1111 & 0.3182 \\ 0.1587 & 1.4286 \end{bmatrix} \] Multiplying this inverse by \( \mathbf{y} \): \[ \mathbf{x} \approx \begin{bmatrix} 19.6721 \\ 38.5246 \end{bmatrix} \]

For Question 5, the input-output matrix \( \mathbf{A} \) is: \[ \mathbf{A} = \begin{bmatrix} 0.2 & 0.3 \\ 0.1 & 0.1 \end{bmatrix} \] and the total output vector \( \mathbf{y} \) is: \[ \mathbf{y} = \begin{bmatrix} 15 \\ 10 \end{bmatrix} \] Calculating \( \mathbf{I} - \mathbf{A} \): \[ \mathbf{I} - \mathbf{A} = \begin{bmatrix} 0.8 & -0.3 \\ -0.1 & 0.9 \end{bmatrix} \] The inverse of this matrix is computed, yielding: \[ (\mathbf{I} - \mathbf{A})^{-1} \approx \begin{bmatrix} 1.25 & 0.4167 \\ 0.1389 & 1.1111 \end{bmatrix} \] Multiplying this inverse by \( \mathbf{y} \): \[ \mathbf{x} \approx \begin{bmatrix} 23.9130 \\ 13.7681 \end{bmatrix} \]

Final Answer